Assuming the question marks are minus signs
to find max, take derivitive and test 0's and endpoints
take derivitive
f'(x)=18x²-18x-108
it equal 0 at x=-2 and 3
if we make a sign chart to find the change of signs
the sign changes from (+) to (-) at x=-2 and from (-) to (+) at x=3
so a reletive max at x=-2 and a reletive min at x=3
test entpoints
f(-3)=83
f(-2)=134
f(3)=-241
f(4)=-190
the min is at x=3 and max is at x=-2
Answer:
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Answer:
ρ_air = 0.15544 kg/m^3
Step-by-step explanation:
Solution:-
- The deflated ball ( no air ) initially weighs:
m1 = 0.615 kg
- The air is pumped into the ball and weight again. The new reading of the ball's weight is:
m2 = 0.624 kg
- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.
m_air = Δm = m2 - m1
m_air = 0.624 - 0.615
m_air = 0.009 kg
- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:
V_air = 4*π*r^3 / 3
V_air = 4*π*0.24^3 / 3
V_air = 0.05790 m^3
- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:
ρ_air = m_air / V_air
ρ_air = 0.009 / 0.05790
Answer: ρ_air = 0.15544 kg/m^3
Answer:
Step-by-step explanation:
<u>Step 1: Set the second equation into the first
</u>
<u>Step 2: Solve the second equation
</u>
Answer:
Answer:
A. Correct: When we plug in g(x) for the x in f(x), we get H(x).
B. Correct: When we plug in g(x) for the x in f(x), we get H(x).
C. Correct: When we plug in g(x) for the x in f(x), we get H(x).
D. Correct: When we plug in g(x) for the x in f(x), we get H(x).
Step-by-step explanation:
<em>Brainliest, please!</em>
